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SOLUTION 7: We are given the function f(x)=2x+x1 and the interval [1,5]. This function is continuous on the closed interval [1,5] since it is the sum of continuous functions y=2x (polynomial) and y=x1 (functional composition of continuous functions y=x1 (polynomial) and y=x (well-known). ). The derivative of f is f(x)=2+(1/2)(x1)1/2=2+12x1 We can now see that f is differentiable on the open interval (1,5). The assumptions of the Mean Value Theorem have now been met. Let's apply the Mean Value Theorem and find all possible values of c in the open interval (1,5). Then f(c)=f(5)f(1)51     2+12c1=(2(5)+51 )(2(1)+11 )4     2+12c1=(10+2)(2+0)4     2+12c1=52     12c1=12     c1=1     (c1 )2=12     c1=1     c=2

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